Respuesta :
Answer: a) 56,550 J b) 30.1 m/s c) 321 m
Explanation:
a) By definition, work, is the process that does an applied force, in order to produce a displacement in the same direction than the applied force, and can be written as follows;
W = F. d. (scalar product of two vectors)
In this case, as the force is parallel to the displacement, work is directly equal to the product of the applied force times the displacement (in magnitude), so we can write the following:
W = F. D = 195 N. 290 m = 56,550 J
b) In absence of friction, the work done by the force is equal to the change in the kinetic energy, as it can be showed using the work-energy theorem.
So, in this case, we can put the following:
W = ΔK ⇒F. D = 1/2 mv²
Solving for v, we get:
v=√2.F.D/m = 30.1 m/s
c) Now, if we assume that there is no friction between the bike and the ground, all the kinetic energy must become gravitational potential energy, at some height h.
We can write the following equation
m.g. h = 1/2 mv²
Simplifying, and taking g = 9.8 m/s², we can find h:
h= 46.2 m
Now, we need to know the distance travelled up the incline, which is related with the height h, by the angle that the incline does with the horizontal, as follows:
sin 8.29° = h /d ⇒ d= h / sin 8.29° = 46.2 m / sin 8.29° = 321 m
A bicyclist must do a work of 5.66 × 10⁴ J to apply a force on the bicycle of 195 N for 290 m and achieve a speed of 30.1 m/s. If the cyclist starts coasting at the bottom of the hill, it travels 320 m.
A bicyclist starts from rest applies a force of 195 N (F) to ride his bicycle across a flat ground for a distance of 290 m (d).
We can calculate the work (w) done by the bicyclist using the following expression.
[tex]w = F \times d \times cos\theta = 195N \times 290m \times cos0\° = 5.66 \times 10^{4} J[/tex]
where,
- θ: angle between F and d
Before the hill, we can state that the work done is equal to the change in the kinetic energy (K), according to the work-energy theorem. Since it starts from rest, the initial kinetic energy is zero.
[tex]w = \Delta K = K_2 - K_1 = K_2 = \frac{1}{2} mv^{2} \\\\v = \sqrt{\frac{2w}{m} } = \sqrt{\frac{2(5.66 \times 10^{4} J )}{125kg} }=30.1 m/s[/tex]
where,
- m: mass
- v: speed
If the cyclist starts coasting at the bottom of the hill, the potential energy (P) at the maximum point will be equal to the kinetic energy at the bottom.
[tex]P_T = K_B = 5.66 \times 10^{4} J = m \times g \times h\\\\h = \frac{5.66 \times 10^{4} J}{m \times g} = \frac{5.66 \times 10^{4} J}{125kg \times 9.81m/s^{2} }= 46.2 m[/tex]
where,
- m: mass
- g: gravity
- h: height
Given the hill has an angle of 8.29 ° with the horizontal, we can calculate the distance (d) traveled using the following trigonometric equation.
[tex]sin8.29\° = \frac{h}{d} \\\\d = 320 m[/tex]
A bicyclist must do a work of 5.66 × 10⁴ J to apply a force on the bicycle of 195 N for 290 m and achieve a speed of 30.1 m/s. If the cyclist starts coasting at the bottom of the hill, it travels 320 m.
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